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(B) The pressure amplitude $\Delta p$ in a sound wave is related to the displacement amplitude $s$ by the formula $\Delta p = B k s$,where $B$ is the

Vedclass · Accepted Answer

$\frac{3}{100} \, mm$

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(B) The pressure amplitude $\Delta p$ in a sound wave is related to the displacement amplitude $s$ by the formula $\Delta p = B k s$,where $B$ is the bulk modulus and $k$ is the wave number.Since $B = ho v^2$ and $k = \frac{\omega}{v} = \frac{2 \pi f}{v}$,we have $\Delta p = (ho v^2) 	imes (\frac{2 \pi f}{v}) 	imes s$.Taking the multiplicative constant to be $1$ (ignoring $2 \pi$),the relation simplifies to $\Delta p = ho v \omega s$.Rearranging for $s$,we get $s = \frac{\Delta p}{ho v \omega} = \frac{\Delta p}{ho v (2 \pi f)}$.Substituting the given values: $s \approx \frac{10}{1 	imes 300 	imes 2 \pi 	imes 1000} \, m$.Ignoring the factor of $2 \pi$ as per the instruction to take the multiplicative constant as $1$: $s \approx \frac{10}{300 	imes 1000} \, m = \frac{1}{30000} \, m$.Converting to millimeters: $s \approx \frac{1}{30000} 	imes 1000 \, mm = \frac{1}{30} \, mm \approx 0.033 \, mm = \frac{3}{100} \, mm$.

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